How do you find the length of a rectangle if its 2 units more than the width and the area is 72

Mathematics

1 answer:

Nastasia [14]3 years ago

4 0

Let width equal X and length equal X+2. Use algebra 72 = X times X+2

You might be interested in

A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one

TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are $\binom44\binom{48}1=48$ possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by $\binom{13}1=13$ . So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

$\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024$

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing $\binom41\binom{48}4=778,320$ possible hands. Exactly 2 aces are drawn in $\binom42\binom{48}3=103,776$ hands. And so on. This gives a total of

$\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656$

possible hands containing at least 1 ace, and hence B occurs with probability

$\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412$

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. $P(A\cap B)=P(A)P(B)$ . This happens if

the hand has 4 aces and 1 non-ace, or
the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So $A\cap B$ consists of 96 possible hands, which occurs with probability

$\dfrac{96}{\binom{52}5}\approx0.0000369$

and so the events A and B are NOT independent.

4 0

2 years ago

When using the binomial formula, which of the following is a correct step in the expansion of (x+y)^4 ?

avanturin [10]

Here is the expansion of (x + y)n for n = 0, 1,…, 5 :

(x + y)0 = 1
(x + y)1 = x + y 
(x + y)2 = x 2 +2xy + y 2 
(x + y)3 = x 3 +3x 2 y + 3xy 2 + y 3 
(x + y)4 = x 4 +4x 3 y + 6x 2 y 2 +4xy 3 + y 4 
(x + y)5 = x 5 +5x 4 y + 10x 3 y 2 +10x 2 y 3 +5xy 4 + y 5

hopefuly this helps you understand

7 0

3 years ago

5(x - 1) and 5x - 1 is it equivalent

suter [353]

No, 5(x-1) expands out to make 5x-5. So 5x-1 and 5x-5 aren’t equivalent.

6 0

2 years ago

Read 2 more answers

I’ll mark you brainlist

kherson [118]

Answer:

Step-by-step explanation: